What is the integral from 0 to b of e^{-x} ?

The integral from 0 to b of e^{-x} is -e^{-b}+1

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integral from 0 to b of e^{-x}

\int_{0}^{b} e^{- x} d x

- e^{- b} + 1

Solution steps

\int_{0}^{b} e^{- x} d x

Apply u-substitution

: \int_{0}^{- b} - e^{u} d u

= \int_{0}^{- b} - e^{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int_{0}^{- b} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - [e^{u}]_{0}^{- b}

Compute the boundaries:

e^{- b} - 1

= - (e^{- b} - 1)

Simplify

= - e^{- b} + 1

\int_{- 2}^{1} x^{2} + 2 x + 1 d x

\int_{- 1}^{2} x + 2 d x

\int_{- 1}^{1} \frac{1}{x ^{2} - 5 x + 6} d x

\int_{2}^{4} \frac{8}{x ^{2}} d x

\int_{1}^{4} x^{- 2} d x

What is the integral from 0 to b of e^{-x} ?
The integral from 0 to b of e^{-x} is -e^{-b}+1